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Example 1: Rectangular Three Point Bending
A rectangular specimen is subjected to a three-point bending test. The specimen is 10 centimeters long, 10 millimeters wide (b) and 10 millimeters tall (h). The specimen is placed on two supports that are 5 cm apart (L), and the actuator is applying a force in the exact middle of the two supports (L/2). Immediately before failure, the Instron records a force (F) of 50N, and a deformation ( ) of 2mm. We need to determine the maximum flexural strength (σ), and Young’s Modulus (E) of the specimen.
To accomplish this task, we are going to use the two following equations:
(Eq. 1.1 and 1.2)
Where M is the moment (or torque) applied at the middle of the specimen, y is the distance from the
center of the specimen to the convex surface, I is the “polar moment of inertia,” a term used to define
how the geometry of the specimen influences its reaction to loads, and m is the slope of the linear portion of the force displacement curve.
First, we must calculate the reaction forces at the supports. We have two unknown values, and therefore must use two equations to solve the system. Based on static mechanics, we can use the following two equations:
∑ (^) (Eq. 1.3)
and
∑ (Eq. 1.4)
In our case, these equations are as follows:
∑
∑
Or
(Eq. 1.5)
Using the (Eq. 1.4), we find:
∑ (^) ( ) ( )
( ) ( )
Solving for we find:
( )
Substituting the value of 25N for back into (Eq. 1.5), we find:
Therefore
Now that we have solved for the reaction forces at the supports, we can calculate the moment acting at the midpoint of the specimen by looking at half of the specimen and using the following equation:
